[Leetcode] Remove Nth Node From End of List (Java)

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

链表操作，多加几个标志指针就比较容易解决了

public class RemoveNthNodeFromEndofList {
static class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode index = head;
ListNode temp = head;
ListNode res = new ListNode(0);
res.next = head;
ListNode pre = res;
while(n!=1) {
temp = temp.next;
n--;
}
while(temp.next!=null) {
pre = index;
index = index.next;
temp = temp.next;
}
pre.next = index.next;
return res.next;
}
public static void main(String[] args) {
ListNode head = new ListNode(1);
//		ListNode l2 = new ListNode(2);
//		ListNode l3 = new ListNode(3);
//		ListNode l4 = new ListNode(4);
//		ListNode l5 = new ListNode(5);
//		head.next=l2;
//		l2.next=l3;
//		l3.next=l4;
//		l4.next=l5;
int n = 1;
ListNode res = new RemoveNthNodeFromEndofList().removeNthFromEnd(head, n);
if(res!=null){
System.out.print(res.val);
res=res.next;
}
while(res!=null){
System.out.print("->"+res.val);
res=res.next;
}
}
}